$ f(x)=\sum_{n=0}^{\infty}{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+1}}}{\left( 2n \right)!}$ $f'''(0)=$
Note that because we are interested in what happens at $x=0$, we only to pay attention to the constant term in the series that represents $f'''$. With that said, we now calculate derivatives. For clarity, we will calculate out one more term than we actually need to do the problem. $\begin{aligned} f'(x)&=1-\dfrac{3{{x}^{2}}}{2!}+\dfrac{5{{x}^{4}}}{4!}-… \\\\ f''x)&=-\dfrac{3\cdot 2x}{2!}+\dfrac{5\cdot 4{{x}^{3}}}{4!}-... \\\\ f'''(x)&=-\dfrac{3\cdot 2}{2!}+\dfrac{5\cdot 4\cdot 3{{x}^{2}}}{4!}-... \end{aligned}$ So $f'''(0)=-3$.